An Additional Mathematics equation containing both log₂x and logₓ2 looks unusual because the unknown x appears as an argument in one logarithm and as the base in the other. The key is change of base: these two logarithms are reciprocals whenever x>0 and x≠1. Set t=log₂x, so logₓ2=1/t, solve a quadratic in t, then convert every admissible t-value back to x.
For example, log₂x+logₓ2=5/2 becomes t+1/t=5/2. Multiplying by 2t yields 2t²−5t+2=0. The roots t=2 and t=1/2 give x=4 and x=√2. Both satisfy the original equation exactly.
Did you know that not every real right-hand side is possible? The reciprocal sum t+1/t is at least 2 when t>0 and at most −2 when t<0. Understanding that range helps students decide whether a logarithmic equation has no solution, one solution or two before reaching for a calculator.
Begin with both real-logarithm domain conditions
The expression log₂x requires x>0. The expression logₓ2 additionally requires its base x to be positive and different from one. Its argument 2 is already positive.
Therefore the common domain of the original equation is x>0 and x≠1. Write those conditions before using change of base or multiplying any fractions.
At x=1, log₂1=0 is defined, but log₁2 is not defined. It is incorrect to include x=1 because one half of the displayed equation happens to have a value there.
The same idea applies to general expressions log_ax+log_xa with a>0 and a≠1. Both logarithms must exist simultaneously; an algebraic transformation cannot create a value at a forbidden base.
Change of base reveals the reciprocal relationship
For valid logarithms, log_bu=ln u/ln b. Apply this to logₓ2 to obtain logₓ2=ln2/ln x.
Meanwhile log₂x=ln x/ln2. Their product is one, provided ln x≠0, which is exactly the restriction x≠1.
Thus logₓ2=1/log₂x. This identity explains why the two apparently different logarithms can be handled using one substituted variable.
Do not try to combine the two terms with a logarithm sum law. The rule log_bM+log_bN=log_b(MN) requires the same base; here the bases are different. Change of base is the correct starting operation.
Choose t equal to the simpler logarithm
Set t=log₂x. Because log₂x=0 exactly when x=1, the original domain implies t≠0.
The reciprocal logarithm becomes logₓ2=1/t. Every permitted x>0 with x≠1 corresponds to a real nonzero t, and conversely x=2^t is positive and not one whenever t≠0.
This correspondence is important: no additional positivity restriction should be imposed on t. Logarithmic outputs may be negative when their positive arguments are below one.
The equation now uses one variable instead of two logarithmic notations. It is an ordinary reciprocal algebra problem whose solutions must still be converted back through x=2^t.
Solve the opening reciprocal equation
Substitute into log₂x+logₓ2=5/2. The transformed equation is t+1/t=5/2, where t≠0.
Multiply by 2t. This multiplication is valid because t cannot be zero. Rearranging gives 2t²−5t+2=0.
Factorise as (2t−1)(t−2)=0. Therefore t=1/2 or t=2. Neither root equals zero, so both satisfy the substituted-variable condition.
Return to the original logarithm: log₂x=1/2 gives x=2^(1/2)=√2. The other root log₂x=2 gives x=4. The complete real solution set is {√2,4}.
Verify the solution x equals four
At x=4, log₂4=2 because 2²=4. The reciprocal-base logarithm is log₄2=1/2 because 4^(1/2)=2.
Their sum is 2+1/2=5/2. Both bases are valid positive numbers different from one and both logarithm arguments are positive.
This check confirms the value in the original equation, not merely in the quadratic after substitution.
A common mistake is to report t=2 as the final answer x=2. But the substitution was t=log₂x, so x=2^t. Keeping the variable names visible through the return step prevents that confusion.
Verify the solution x equals square root two
At x=√2, log₂√2=1/2 because √2=2^(1/2).
The reciprocal logarithm is log_(√2)2=2 because (√2)²=2. Their sum is 1/2+2=5/2.
Thus the second candidate also satisfies the equation and its domain. There is no need to convert √2 into a long decimal.
The two answers are connected by the transformed t-roots 2 and 1/2, whose product is one. They are not arbitrary guesses found by trial. The algebraic structure guarantees the pair.
Why negative t-values can be valid
If 0 The corresponding reciprocal-base logarithm log_(1/4)2 equals −1/2 because (1/4)^(−1/2)=2. Both logarithms are defined. Their sum is negative, not invalid. Therefore the substituted variable t may be any real number except zero. Rejecting negative t-roots merely because logarithm inputs must be positive would confuse the logarithm’s argument with its output. A careful solver states the original x-domain and the transformed t-domain separately. Solve log₂x+logₓ2=−5/2. After the same substitution, t+1/t=−5/2. Multiply through by 2t and rearrange to 2t²+5t+2=0, or (2t+1)(t+2)=0. Thus t=−1/2 or t=−2. Both are nonzero and valid. Returning through x=2^t gives x=2^(−1/2)=1/√2 and x=2^(−2)=1/4. Both x-values lie strictly between zero and one. This is consistent with both logarithms being negative when base 2 is used and the variable x lies below one. When t>0, the square (t−1)² is non-negative. Expand to t²−2t+1≥0. Divide by positive t to obtain t−2+1/t≥0, hence t+1/t≥2. Equality occurs exactly when t=1. In the original variable, log₂x=1 gives x=2. Therefore the logarithmic expression cannot equal 1 or 3/2 when x>1. Its smallest value on that branch is exactly two, attained at x=2. This algebraic inequality is stronger than checking a few numerical examples because it applies to every positive t, however small or large. For t<0, write u=−t>0. Then t+1/t=−u−1/u. Since u+1/u≥2 for every positive u, its negative satisfies −u−1/u≤−2. Equality occurs when u=1, so t=−1 and x=2^(−1)=1/2. Thus on the branch 0 The proof avoids an unsafe direct multiplication by a negative t. It also explains why the positive and negative branches have separate ranges with a missing interval between −2 and 2. For a real parameter K, consider log₂x+logₓ2=K. If K>2, there are two valid real x-solutions, both greater than one. At K=2, there is one solution x=2. If −2 This classification follows from the range of t+1/t on its two nonzero branches. It is independent of a particular quadratic’s convenient factorisation and helps detect impossible answers immediately. Write t+1/t=K and remember t≠0. Multiply by t to obtain t²+1=Kt. Rearrange to t²−Kt+1=0. The discriminant is Δ=K²−4. The quadratic formula gives t=[K±√(K²−4)]/2 when Δ≥0. The two roots multiply to one, so neither can equal zero. However, the discriminant must be interpreted with the sign of K. For |K|≥2, the transformed roots have the same sign as K, and both are admissible nonzero real logarithm outputs. If |K|<2, the roots are nonreal and no real x-solutions exist. For |K|>2, define t_+=[K+√(K²−4)]/2 and t_-=[K−√(K²−4)]/2. The corresponding original values are x_+=2^(t_+) and x_-=2^(t_-). Both are positive and not one. At K=2 or K=−2, the two t-expressions coincide, producing one distinct x-solution rather than two. The exact exponential forms may be preferable to decimals when the right-hand side does not yield simple factors. If rounded values are requested, calculate them at the final step and check the original logarithm domain before reporting them. At K=2, the transformed quadratic becomes t²−2t+1=(t−1)²=0. The only distinct t-value is 1, so the only original solution is x=2. At K=−2, the transformed quadratic is (t+1)²=0, giving x=1/2. The repeated algebraic root does not represent two different logarithmic inputs. These boundary cases correspond to turning points of the reciprocal sum on its separate positive and negative branches. Counting roots with multiplicity is not the same as counting distinct solutions to the original equation. If K=0, the equation becomes t+1/t=0 with t≠0. Multiplying by t gives t²+1=0, or t²=−1. There is no real t satisfying this condition. The range argument provides the same conclusion: t+1/t is never between −2 and 2, so zero is impossible. This is a useful check on the transformed quadratic. A student should not invent logarithms of imaginary numbers when the question asks for real x. The correct answer is simply no real solution. For base 2, which is greater than one, log₂x is positive precisely when x>1. Therefore the positive t-branch corresponds to the original interval (1,∞). Similarly, log₂x is negative precisely when 0 The missing point x=1 separates the two branches and is not permitted as a logarithm base. Near x=1, the reciprocal 1/t becomes very large in magnitude. Keeping this correspondence clear is especially useful when solving inequalities or sketching the expression as a function of x. The sign of the right-hand target immediately tells us which x-branch might contain solutions. The function F(t)=t+1/t is not monotonic on the whole positive half-line. It decreases on 0 Thus for any K>2, two different positive t-values produce the same output. They are reciprocals. The same phenomenon occurs on the negative half-line: for K<−2, two negative reciprocal values produce the same sum. This explains why a single logarithmic equation can have two distinct x-solutions on one side of x=1. The logarithm log₂x itself is one-to-one, but the subsequent reciprocal sum is not. Differentiate F(t)=t+1/t for t≠0. The derivative is F′(t)=1−1/t². It vanishes at t=1 and t=−1. On the positive branch, F decreases for 0 On the negative branch, the derivative is positive when |t|>1 and negative when |t|<1. This gives a maximum at t=−1, with value −2. The calculus classification matches the elementary square-inequality proof. Both approaches show the forbidden output interval (−2,2) and provide an independent way to check the solution count. Solve log₂x+logₓ2≥5/2. Substitute t=log₂x≠0, so t+1/t≥5/2. No negative t can satisfy this because the negative branch is at most −2. Therefore restrict to t>0, where multiplying by 2t preserves the inequality direction. We obtain 2t²−5t+2≥0, or (2t−1)(t−2)≥0. Hence 0 Because x=2^t increases with t, the result is 1 The inequality t+1/t≥5/2 cannot safely be multiplied by t before knowing its sign. Multiplication by a negative number would reverse an inequality. The range analysis first shows that any solution must have t>0. Only then is multiplying by t direction-preserving. For equality problems, multiplying by a known nonzero t does not introduce an inequality-direction issue. For inequalities, the sign is essential. This is a general A-Math lesson about rational inequalities. A familiar algebraic step may require a different logical treatment when an equal sign is replaced by <, >, ≤ or ≥. Consider log₂x+logₓ2≤−5/2. Any solution must be on the negative t-branch, so t<0. The reciprocal sum is at most −5/2 when t≤−2 or −1/2≤t<0. This can be checked by testing values around the equality roots t=−2 and t=−1/2. Convert back using increasing x=2^t. The first region gives 0 Thus the complete solution is (0,1/4]∪[1/√2,1). The endpoint x=1 is excluded because the variable-base logarithm is undefined there. Let a>0, a≠1, and consider log_ax+log_xa=K. Set t=log_ax, with x>0 and x≠1. By change of base, log_xa=1/t. The transformed equation is again t²−Kt+1=0. Thus the possible t-values depend on K but not on the particular fixed valid base a. Returning to x uses x=a^t, which does depend on the base. This shows which part of the problem is universal: reciprocal-logarithm algebra. The numerical x-solutions are produced by the final exponential mapping and therefore vary with the fixed base. Suppose the fixed base is a=1/2. Then log_(1/2)x is positive when 0 The same positive target K=5/2 gives t=1/2 or 2. Returning through x=(1/2)^t produces x=1/√2 or 1/4, both below one. The range in K remains (−∞,−2]∪[2,∞), but the placement of its positive and negative branches in the x-domain is reversed. A student should not automatically infer x>1 from positive t unless the fixed logarithm base is greater than one. Base monotonicity affects the final interpretation. The usual logarithm product rule log_bM+log_bN=log_b(MN) is valid for the same base b and positive arguments. It does not allow log₂x+logₓ2 to be written as log₂(2x) or log_x(2x). Such a rewrite silently changes one of the logarithm bases. Likewise, the reciprocal identity does not say the two logarithms add to one. It says their product equals one. To check, at x=4 the two logarithms are 2 and 1/2. Their product is one, but their sum is 5/2. This simple counterexample catches the most common false rule. The notation logₓ2 means the logarithm of the number two using x as the base. It is not the reciprocal of log₂x by definition; the reciprocal relationship follows from change of base under the correct domain. An inverse logarithmic function means the exponential function that reverses a fixed-base logarithm, such as 2^t reversing log₂x. These ideas are connected but not identical. The original task asks for x such that a sum of two log values reaches a target, not for the inverse function of their sum. Keeping the language precise helps students choose change of base, substitution and exponentiation in the correct order. For the fixed base two, define G(x)=log₂x+logₓ2 on x>0, x≠1. On 0 On x>1, the graph’s values are at least 2, reaching 2 at x=2. As x approaches one from the right or grows without bound, the output becomes arbitrarily positive. Thus no horizontal line y=K with −2 The equation can also be read in reverse. If x=8, then log₂8=3 and log₈2=1/3. Therefore G(8)=3+1/3=10/3. The partner transformed t-value is 1/3, which corresponds to x=2^(1/3). Thus both x=8 and x=∛2 solve the equation G(x)=10/3. This offers a way to construct exact practice questions with two known answers. The partner x-values need not multiply to a fixed number. It is their logarithm outputs t and 1/t that multiply to one, not the original x inputs. Check the very first line: did the learner state x>0 and x≠1? Then check whether change of base was used correctly to obtain one reciprocal quantity. Next ask whether the substitution t=log₂x was kept distinct from x. In an equality, ensure every nonzero t-root was retained and converted back through x=2^t. For inequalities, verify the sign of t before multiplication and the monotonic direction of the chosen exponential base. Finally substitute every x-candidate into the original pair of logarithms. These targeted checks identify domain, algebra, sign and return-substitution errors separately instead of treating all wrong answers as one generic issue. Question one: solve log₂x+logₓ2=5/2. Answer: x=√2 or x=4. Question two: solve log₂x+logₓ2=−5/2. Answer: x=1/4 or x=1/√2. Question three: solve log₂x+logₓ2=2. Answer: x=2 only. Question four: solve log₂x+logₓ2=−2. Answer: x=1/2 only. Question five: solve log₂x+logₓ2=0. Answer: no real solution. Question six: find K for which log₂x+logₓ2=K has real solutions. Answer: K≤−2 or K≥2. Question seven: solve log₂x+logₓ2≥5/2. Answer: 1 For each, begin with x>0, x≠1, then use t=log₂x. For the target K=5/2, the substituted logarithm values are t=2 and t=1/2. Their product is one, so they are reciprocals. But the original x-values are x=2²=4 and x=2^(1/2)=√2. Their product is 4√2, not one. The reciprocal relationship is between the values of the two logarithms, or between the two roots of the auxiliary quadratic in t. It does not mean that the original x-values are reciprocal. This distinction helps when checking answers. A student who takes x=4 and automatically writes x=1/4 as its partner has confused the substituted level with the original exponential mapping. The correct second answer is √2. Let G(x)=log₂x+logₓ2 on x>0, x≠1. Put t=log₂x, so G=t+1/t and dt/dx=1/(x ln2). By the chain rule, G′(x)=(1−1/t²)/(x ln2). The denominator x ln2 is positive. Therefore the sign of G′ depends on 1−1/t². Its zeros occur when t²=1, giving t=−1 or t=1, hence x=1/2 or x=2. The first is the local maximum −2 on the branch 0 Consider the related equation log₂x+4logₓ2=5. Use the same domain x>0, x≠1, and let t=log₂x≠0. Then t+4/t=5. Multiply by t to get t²−5t+4=0, or (t−1)(t−4)=0. Thus t=1 or t=4, giving x=2 or x=16. Check directly: at x=2 the logarithms are 1 and 1, so 1+4(1)=5. At x=16 they are 4 and 1/4, so 4+4(1/4)=5. The coefficient four changes the reciprocal sum but not the underlying substitution method. It also shows why coefficients should remain attached to the correct logarithm term during change of base. For c>0, consider H(t)=t+c/t with t≠0. On t>0, the inequality (t−√c)²≥0 gives t+c/t≥2√c. On t<0, put u=−t>0 to obtain H(t)=−(u+c/u)≤−2√c. Thus the range is (−∞,−2√c]∪[2√c,∞), and equality is attained at t=±√c. The ordinary reciprocal-logarithm equation corresponds to c=1, with thresholds ±2. The weighted example log₂x+4logₓ2 has c=4, so its target must lie at or above 4 or at or below −4. Its right-hand side five is therefore feasible. Suppose log₂x+logₓ2+c=5/2, where c is a real parameter. Rearrange to log₂x+logₓ2=5/2−c. The reciprocal-logarithm range requires 5/2−c≥2 or 5/2−c≤−2. The first condition gives c≤1/2. The second gives c≥9/2. Therefore the equation has at least one real solution precisely when c≤1/2 or c≥9/2. At the two boundary values there is one solution; farther outside them there are two. No logarithmic calculation in x is needed to classify the parameter. Solve log₄x+logₓ4=5/2. Let t=log₄x. The same quadratic gives t=2 or t=1/2. Convert using the correct fixed base: x=4²=16 or x=4^(1/2)=2. These differ from the base-two answers 4 and √2, even though the transformed t-values and right-hand target are identical. This is a useful last-step check. A solver may correctly identify t=2 or 1/2 but then exponentiate with base two out of habit. The base used to define t determines the final conversion and must be preserved until the end. This topic brings together change of base, logarithmic domains, reciprocal algebra, quadratics, functions and inequalities. Variable-base logarithms and range classifications can be enrichment depending on the student’s particular Additional Mathematics syllabus and school sequence. Singapore learners can consult the relevant SEAB references for 2026 O-Level Additional Mathematics or 2027 SEC G3. Related eduKateSG guides include How to Spot a Hidden Quadratic in a Logarithmic Equation, Why a Logarithmic Inequality Reverses When Its Base Is Below One and How to Solve an Exponential Equation with Its Reciprocal. Browse the Additional Mathematics archive, read the eduKate reference tutorial format or enquire about suitable small-group Mathematics lessons. The goal is to recognise structure without losing the real logarithm domain. Properly taught kids shine a bright light into the future.A negative right-hand side produces fractional x-values
Prove the positive reciprocal sum is at least two
Prove the negative reciprocal sum is at most minus two
Classify every possible target K
Derive the general quadratic in t
Write the general x-solutions exactly
Why discriminant zero means one x-solution
Why K equal zero has no solution
Understand which original x-branch corresponds to each sign
The expression is not one-to-one on either branch
Use differentiation to check the range
An exact inequality example
Why the positivity case split matters for inequalities
Solve a negative-target inequality carefully
A changed fixed base retains the reciprocal structure
A base below one reverses the x-branch interpretation
Do not split or combine unlike logarithm bases
Distinguish a reciprocal logarithm from an inverse function
A graph helps explain the missing middle outputs
Recover a target from a given value of x
A first-error checklist for student work
Practice questions with exact answers
Why the two original x-solutions are not reciprocal numbers
Recover the derivative of the original logarithmic sum
A weighted reciprocal-logarithm example
A general weighted logarithmic range
A parameter on the right can be classified before solving
A changed base can alter numerical answers without changing the target threshold
Syllabus fit and further reading